What is the extreme f(x)=x^{4/3}-2x^{1/3} ?

The extreme f(x)=x^{4/3}-2x^{1/3} is Minimum(1/2 ,(1/2)^{4/3}-4^{1/3})

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extreme f(x)=x^{4/3}-2x^{1/3}

extreme points

f (x) = x^{\frac{4}{3}} - 2 x^{\frac{1}{3}}

Minimum (\frac{1}{2}, (\frac{1}{2})^{\frac{4}{3}} - 4^{\frac{1}{3}})

Solution steps

f^{'} (x) = \frac{4 x - 2}{3 x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = \frac{1}{2}

into

x^{\frac{4}{3}} - 2 x^{\frac{1}{3}} : (\frac{1}{2})^{\frac{4}{3}} - 4^{\frac{1}{3}}

Minimum (\frac{1}{2}, (\frac{1}{2})^{\frac{4}{3}} - 4^{\frac{1}{3}})

\frac{x}{x ^{2} + 12 x + 35}

f (x, y) = 2 x y - x^{2} - 2 y^{2} + 3 x + 4

f (x) = 2 x^{3} - 6 x + 1

f (x) = 2 x^{3} - 6 x + 5

\frac{7 x}{x ^{2} - 9}

What is the extreme f(x)=x^{4/3}-2x^{1/3} ?
The extreme f(x)=x^{4/3}-2x^{1/3} is Minimum(1/2 ,(1/2)^{4/3}-4^{1/3})